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A380649 Rectangular array ((-1)*D(i,j,1,2)) read by descending antidiagonals: D(i,j,s,n) denotes the determinant of the matrix described in Comments.

%I A380649 #6 Feb 03 2025 21:35:27
%S A380649 1,4,3,8,7,6,13,12,11,10,19,18,17,16,15,26,25,24,23,22,21,34,33,32,31,
%T A380649 30,29,28,43,42,41,40,39,38,37,36,53,52,51,50,49,48,47,46,45,64,63,62,
%U A380649 61,60,59,58,57,56,55,76,75,74,73,72,71,70,69,68,67,66
%N A380649 Rectangular array ((-1)*D(i,j,1,2)) read by descending antidiagonals: D(i,j,s,n) denotes the determinant of the matrix described in Comments.
%C A380649 Suppose that (m(i,j)) is a rectangular array of infinitely many rows and infinitely many columns. For integers s>=1 and n>=1, let M(i,j,s,n) be the nXn matrix (m(i+h*s,j+k*s)), where h=0..n-1, k=0..n-1.
%C A380649 Let D(i,j,s,n) and P(i,j,s,n) denote the determinant and permanent of M(i,j,s,n), respectively. For A380649, we take (m(i,j)) to be the natural number array (see A000027, A185787, and A144112), and ((-1)*D(i,j,1,2)) is as shown below in Example.
%C A380649 * D(i,j,1,1) = M(i,j,1,1) = m(i,j) has linearly recurrent row sequences, all with signature (3,-3,1).
%C A380649 * D(i,j,1,2) has linearly recurrent row sequences, all with signature (3,-3,1).
%C A380649 * (-1)*D(i,j,s,3) is the constant array in which every term is s^6, for all i,j,s.
%C A380649 * D(i,j,s,n) is the constant 0 array for all n>=4, for all i,j,s.
%C A380649 * P(i,j,s,n) depends only on n, and the rows all have the following linear recurrence signature:
%C A380649 b(2n+1,1), - b(2n+1,2), b(2n+1-3),..., -(2n+1,2n), 1, where b=binomial.
%C A380649 ((-1)*D(i,j,1,2)) includes, exactly once, every positive integer not in A000096. The order array of ((-1)*D(i,j,1,2)) is the array in Example in A038722; see A333029 for the definition of order array.
%e A380649 Corner of (-1)*D(i,j,1,2):
%e A380649    1   4    8   13   19   26   34   43   53   64   76   89
%e A380649    3   7   12   18   25   33   42   52   63   75   88  102
%e A380649    6  11   17   24   32   41   51   62   74   87  101  116
%e A380649   10  16   23   31   40   50   61   73   86  100  115  131
%e A380649   15  22   30   39   49   60   72   85   99  114  130  147
%e A380649   21  29   38   48   59   71   84   98  113  129  146  164
%e A380649   28  37   47   58   70   83   97  112  128  145  163  182
%e A380649   36  46   57   69   82   96  111  127  144  162  181  201
%e A380649   45  56   68   81   95  110  126  143  161  180  200  221
%e A380649   55  67   80   94  109  125  142  160  179  199  220  242
%e A380649   66  79   93  108  124  141  159  178  198  219  241  264
%e A380649   78  92  107  123  140  158  177  197  218  240  263  287
%e A380649 m(1,1) = 1, so M(1,1,1,2) is the matrix having (row 1) = (1,2) and (row 2) = (3,5), so (-1)*D(1,1,1,2) = -(1*5-2*3) = 1.
%t A380649 s = 1; n = 2; z = 12;
%t A380649 r[n_, k_] := n + (n + k - 2)*(n + k - 1)/2;
%t A380649 Grid[Table[r[n, k], {n, 1, z}, {k, 1, z}]]  (* Array A000027 *)
%t A380649 FindLinearRecurrence[Table[r[1, k], {k, 1, 20}]]
%t A380649 t[i_, j_] := Table[r[i, j + k*s], {k, 0, n - 1}];
%t A380649 d[i_, j_] := -Det[Table[t[i + k*s, j], {k, 0, n - 1}]];  (* (-1)*D(i,j,s,n) *)
%t A380649 Grid[Table[d[i, j], {i, 1, z}, {j, 1, z}]]  (* array *)
%t A380649 FindLinearRecurrence[Table[d[12, k], {k, 1, 20}]]
%t A380649 Table[d[k, m - k], {m, 2, z}, {k, 1, m - 1}] // Flatten (* sequence *)
%Y A380649 Cf. A000027, A000096, A038722, A185787, A333029, A380660, A380661.
%K A380649 nonn,tabl
%O A380649 1,2
%A A380649 _Clark Kimberling_, Jan 31 2025